1. Mathematical Background: Extension Fields
Network Security
Design Fundamentals
ET-IDA-082
Tutorial-4
Mathematical Background: Extension Fields
, v24
Prof. W. Adi

2. Irreducible Polynomials and extension Fields GF(2m)
g(x) is called Irreducible Polynomial of degree m over GF(2)
where ai  GF(2) and Factorization is not possible over GF(2)
g(x) =
The period e of g(x) is the smallest e such that xe = [mod g(x)]
e is actually the order of x modulo g(x). e divides 2m -1
If e = 2m -1 the the polynomial is called primitive
The reciprocal polynomial is defined as g*(x) = xn g(1/x)
The period of the reciprocal polynomial g*(x) is equal to that of g(x)
GF (2m )
The ring of polynomials Zg(x) modulo an irreducible polynomial of degree m
over GF(2) is an extension field with 2m elements
The order of any element in GF(2m) is a divisor of 2m -1 (Lagrange theorem)

3. Problem 4-1: Polynomials over a field
Give the corresponding vector representation of the polynomials
1 + x2 + x5 + x9
1 + x3 + x8 + x12
x2 + x5 + x7
x + x9 + x10
1 + x8
1 + x + x2 + x3+ x4 + x5 + x6 + x7
Solution 4-1:
LSB MSB
1 + x2 + x5 + x =
1 + x3 + x8 + x =
x2 + x5 + x =
x + x9 + x =
1 + x =
1 + x + x2 + x3+ x4 + x5 + x6 + x7 =

4. Problem 4-2: Polynomials over a field
Compute the following polynomial products over GF(2)
(1 + x2 ) (1 + x2 + x5 )
(x3 + x4)(1 + x3 + x8 + x12)
X3 (1 + x3 + x5 + x6)
Compute the following polynomial products over GF(3)
(1 + 2x2 ) (1 + x2 +2x5 )
(x + 2x3)(1 + 2x3 + x8 + x12)
Compute the following polynomial products over GF(7)
(2 + 4x2 ) (1 + 3x2 + 5x5 )
(3x + 5x3)(6 + 2x3 + 3x8)
Solution 4-2:
products over GF(2)
(1 + x2 ) (1 + x2 + x5 ) = 1 + x2 + x5 + x2 + x4 + x7 = x2 + x4 + x5 + x7 = 1 + x4 + x5 + x7
(x3 + x4)(1 + x3 + x8 + x12) = x3 + x6 + x11 + x x4 + x7 + x12 + x16 = x3 + x4 + x6 + x7 + x11 + x12 + x15 + x16
X3 (1 + x3 + x5 + x6) = X3 + x6 + x8 + x9

5. Solution 4-2 cont.: products over GF(3)
(1 + 2x2 ) (1 + x2 +2x5 ) = 1 + x2 +2x x2 + 2x4 +4x7 = 1 + 3x x4 +2x5 +4x7 = 1 + 2x4 +2x5 +1x7
(x + 2x3)(1 + 2x3 + x8 + x12) =
Products over GF(7)
(2 + 4x2 ) (1 + 3x2 + 5x5 ) = 2 + 6x2 +10x5 + 4x2 + 12x4 + 20x7 = x2 +12x4 + 10x x7
= 2 +3x2 + 5x4 + 3x5 + 6x7
(3x2 + 5x3)(6x2 + 2x3 + 3x8) =

6. Problem 4-3: Polynomials division over a finite field
Compute the following polynomial divisions over GF(2)
(x12 + x8 + x6 + x2 + 1)  (x5+ x2 +1 )
Solution 4-3:
( x12 + x8 + x6 + x2 + 1)  ( x5+ x2 +1 ) -
(x12 + x9 + x7) x7
+ x4
+ x3
+ x2
+ 1
x9 + x8 + x7 + x6+ x2 + 1
Q/x)
in GF(2):
1+1=2=0
=> 1=-1
=>Addition
Is equal to
subtraction
x9 + x6 + x4
x8 + x7 + x4 + x2 + 1
x8 + x5 + x3
x7 + x5 + x4 + x3 + x2 + 1
x7 + x4 + x2
Remainder of the division
R/x)
x5 + x
x5 + x
x3 + x2

7. Solution: Compute gcd [ P1(x) , P2(x) ] = A(x) P1(x) + B(x) P2(x)
Problem 4.4: Find the multiplicative inverse of x + 1 modulo x5 + x3 + 1
Solution: Compute gcd [ P1(x) , P2(x) ] = A(x) P1(x) + B(x) P2(x)
if gcd =1, then the inverse is B(x)
Extended gcd Algorithm:
A2 = A1 – q A2
B2 = B1 – q B2
P1(x)
P2(x)
A1(x)
A2(x)
B1(x)
B2(x)
Qx)
R(x)
x5 + x3 + 1
x + 1 1 1
x4 + x3 1
0 – (x4 + x3 ). 1=
x4 + x3
x + 1 1
x +1 1 1
=> (x4 + x3 )  (x + 1 ) modulo (x5 + x3 + 1)
x5 + x3 + 1=0
x5 = x3 + 1
Check: (x + 1) (x4 + x3 ) = x5+ x4 + x4 + x3  1 modulo (x5 + x3 + 1)

8. Problem 4-5: Elements order over an extension field
Compute the exponents of the element x over GF(25) which is generated by the irreducible polynomial
P(x)= (x5+ x2 +1 )
Solution 4-5:
If P(x)= x5+ x2 +1 is the modulus then it is equal to zero
That is x5+ x2 +1 = 0 Thus x5 = x2 +1
Let us compute the exponents of x over this field:
x = x mod (x5+ x2 +1 )
x2= x mod (x5+ x2 +1 )
x3= x mod (x5+ x2 +1 )
x4= x mod (x5+ x2 +1 )
x5= x mod (x5+ x2 +1 )
x6= x (x2 +1)= x3 +x mod (x5+ x2 +1 )
x7= x (x3 +x) = ( x4 +x2 ) mod (x5+ x2 +1 )
x8= x5 + x3 = x x3 mod (x5+ x2 +1 )
x9= x4 + x3 + x mod (x5+ x2 +1 ) ….
x14 = (x7)2 = ( x4 +x2 )2 = x8 + x4 = x3 + x x4
x15 = ( x3 + x x4 ) x = x4 + x3 +x + x5 = x4 + x3 +x + x2 +1
x16 = (x8)2 = ( x3 +x2 +1 )2 = x6 + x4 +1= x3 + x + x4 +1 =
Important notice:
In GF (25): the order of any element
Is a divisor of 25-1=31
Divisors of 31 are 1 and 31 !
=> The order can be either 1 or 31 !

9. Problem 4-6: Arithmetic in GF(25)
Given GF(25) generated by the irreducible polynomial g(x) = x5 + x3 + 1
Design a circuit which multiplies any serial data stream I(x) by the inverse of the polynomial p(x)= x + 1 in GF(25). Check the circuit by multiplying (1 + x2 + x3 ) *(x + 1)-1
Solution 4-6: Compute gcd [ g(x) , p(x) ] = A(x) g(x) + B(x) p(x)
if gcd =1, then the multiplicative inverse of p(x) is B(x)
Extended gcd Algorithm:
A2 = A1 – q A2
B2 = B1 – q B2
P1(x)
P2(x)
A1(x)
A2(x)
B1(x)
B2(x)
Qx)
R(x)
x5 + x3 + 1
x + 1 1 1
x4 + x3 1
0 – (x4 + x3 ). 1
= (x4 + x3 )
x + 1 1
x +1 1 1
=> (x + 1 ) -1  (x4 + x3 ) modulo (x5 + x3 + 1)
x5 + x3 + 1=0
x5 = x3 + 1
Check: (x + 1) (x4 + x3 ) = x5+ x4 + x4 + x3  1 modulo (x5 + x3 + 1)

10. A circuit which multiplies any serial data stream I(x) by the inverse of the polynomial p(x)= x + 1 that is p(x)-1 = x3 + x4 modulo the irreducible polynomial g(x) = x5 + x3 + 1
(x + 1 )  (x4 + x3 )
LSB .. MSB
I(x)= 1 + x2 + x3 =
Remainder
MSB
Check:
(1 + x2 + x3)(x4 + x3 )
= x4 + x6 +x7 +x3 + x5 +x6
= x4 + x7 +x3 + x5
= x4 + x2 +1+x3 +x x3
= x4 + x2 +x3
x5 = 1 +x3
x6 = x +x4
x7 = x2 +x5= x2 +1+x3
Remainder
LSB x3 x4 r0 r1 r2 + r3 + r4 1 x3 x5
Clocking in the data stream 1011
States before clocking each input
Input register state
Input register state
Input register state
Input register state
Input x register state
g(x) = x5 + x3 + 1
LSB .. MSB
Result of multiplication= x2 + x3 +x4

11. Problem 4-7: ad-hoc Class exercise
Select a polynomial as a modulus for GF(28)
Compute a primitive element
Wich are the possible multiplicative orders in GF(28)
How many elements do exist from each possible order
Compute other 5 primitive elements
Compute one element for each possible multiplicative order
Solution 4-7: